Calculus 10th Edition

by
Larson, Ron; Edwards, Bruce H.

Published by
Brooks Cole

ISBN 10:
1-28505-709-0

ISBN 13:
978-1-28505-709-5

Chapter 6 - Differential Equations - 6.3 Exercises: 25

Answer

$$y=\sqrt{\frac{x^2}{4}+4}$$

Work Step by Step

$y'=\frac{x}{4y}$
$4yy'=x$
$\int 4ydy=\int xdx$
$2y^2=\frac{x^2}{2}+C$
$y=\sqrt{\frac{x^2}{4}+C}$
Now we need this equation to pass through $(0,2)$ so we apply it as an initial condition:
$2=\sqrt{0+C}$
$C=4$
So our equation becomes
$$y=\sqrt{\frac{x^2}{4}+4}$$